If a snowball melts so that its surface area decreases at a rate of 7 cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 10 cm. (Round your answer to three decimal places.)
Please try again. Keep in mind that the surface area of a snowball (sphere) with radius 
r
 is 
A = 4𝜋r2.
 Differentiate this equation with respect to time, 
t,
 using the Chain Rule, to find the equation for the rate at which the area is decreasing, 
dA
dt
.
 Then, use the values from the exercise to evaluate the rate of change of the radius of the sphere. Have in mind that the diameter is twice the radius.

To solve this problem, we are given the rate of change of the surface area of a snowball and asked to find the rate at which the diameter is decreasing when the diameter is 10 cm.

### Step 1: Surface area of a sphere
The surface area \( A \) of a sphere with radius \( r \) is given by:

\[
A = 4\pi r^2
\]

### Step 2: Differentiate with respect to time
We differentiate both sides of this equation with respect to time \( t \), applying the Chain Rule:

\[
\frac{dA}{dt} = \frac{d}{dt}(4\pi r^2) = 8\pi r \frac{dr}{dt}
\]

This gives us the relationship between the rate of change of surface area \( \frac{dA}{dt} \) and the rate of change of the radius \( \frac{dr}{dt} \).

### Step 3: Plug in known values
We are told that the surface area is decreasing at a rate of 7 cm²/min, so:

\[
\frac{dA}{dt} = -7 \, \text{cm}^2/\text{min}
\]

The diameter is given as 10 cm, so the radius \( r \) is:

\[
r = \frac{10}{2} = 5 \, \text{cm}
\]

Substitute these values into the differentiated equation:

\[
-7 = 8\pi (5) \frac{dr}{dt}
\]

### Step 4: Solve for \( \frac{dr}{dt} \)
Now, solve for \( \frac{dr}{dt} \) (the rate at which the radius is changing):

\[
\frac{dr}{dt} = \frac{-7}{8\pi (5)} = \frac{-7}{40\pi}
\]

Using \( \pi \approx 3.1416 \), we compute:

\[
\frac{dr}{dt} = \frac{-7}{40 \times 3.1416} \approx \frac{-7}{125.664} \approx -0.0557 \, \text{cm/min}
\]

### Step 5: Find the rate of change of the diameter
Since the diameter is twice the radius, the rate of change of the diameter \( \frac{dD}{dt} \) is:

\[
\frac{dD}{dt} = 2 \times \frac{dr}{dt} = 2 \times (-0.0557) \approx -0.111 \, \text{cm/min}
\]

### Final Answer:
The diameter of the snowball is decreasing at approximately \( 0.111 \, \text{cm/min} \) when the diameter is 10 cm.

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Do you need any further clarification or details?

Here are 5 related questions for further exploration:
1. How would the rate of decrease of the diameter change if the diameter were different?
2. What is the formula for the volume of a sphere, and how does the rate of change of the volume relate to this problem?
3. What happens to the rate of change of the diameter as the snowball shrinks?
4. How can we interpret the negative sign in the rate of change in a physical context?
5. How would this problem change if the rate of surface area decrease were not constant?

**Tip**: When solving related rates problems, always keep track of units and signs, as they indicate whether a quantity is increasing or decreasing.

If a snowball melts so that its surface area decreases at a rate of 7 cm²/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 10 cm. (Round your answer to three decimal places.)

In this calculus problem, learn how to calculate the rate at which the diameter of a snowball decreases when its surface area decreases at a rate of 7 cm²/min. Using the surface area formula of a sphere and related rates concepts, this problem is suitable for Grades 11-12 students studying calculus.

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